Radius of Uniform Convexity of some special functions

Theorem 1.1. Let f(z) is of the form (1.1). Then, f is a uniformly convex functions if and only if $$\Re\left(1+\frac{zf''(z)}{f'(z)}\right) > \left|\frac{zf''(z)}{f'(z)}\right|, z \in \mathbb{D}.$$ On the other hand, the concept of the radius of uniform convexity is defined by (see [15]) <span id="page-1-1"></span> $$r^{uc}(f) = \sup \left\{ r \in (0, r_f) : \Re\left(1 + \frac{zf''(z)}{f'(z)}\right) > \left|\frac{zf''(z)}{f'(z)}\right|, z \in \mathbb{D} \right\}.$$ Thanks to above theorem we can determine the radius of uniform convexity for the functions of the form (1.1). Also, we will need the following lemma in the sequel.

Lemma 1.1. ([15]). If $a > b > r \ge |z|$ , and $\lambda \in [0, 1]$ , then $$\left| \frac{z}{b-z} - \lambda \frac{z}{a-z} \right| \le \frac{r}{b-r} - \lambda \frac{r}{a-r}.$$ The followings are very simple consequences of this inequality <span id="page-1-0"></span>(1.3) $$\Re\left(\frac{z}{b-z} - \lambda \frac{z}{a-z}\right) \le \frac{r}{b-r} - \lambda \frac{r}{a-r}$$ and (1.4) $$\Re\left(\frac{z}{b-z}\right) \le \left|\frac{z}{b-z}\right| \le \frac{r}{b-r}.$$

Theorem 2.1. Let $\nu > -1$ , $s \in \{2,3\}$ and $q \in (0,1)$ . The following assertions are true. a. Suppose that $\nu > 0$ . Then, the radius of uniform convexity of the function $z \mapsto f_{\nu}^{(s)}(z;q)$ is the smallest positive root of the equation $$1 + 2r \frac{\left(f_{\nu}^{(s)}(r;q)\right)''}{\left(f_{\nu}^{(s)}(r;q)\right)'} = 0.$$ b. The radius of uniform convexity of the function $z \mapsto g_{\nu}^{(s)}(z;q)$ is the smallest positive root of the equation $$(2\nu - 1)(\nu - 1)J_{\nu}^{(s)}(r;q) + (5 - 4\nu)r\left(J_{\nu}^{(s)}(r;q)\right)' + 2r^2\left(J_{\nu}^{(s)}(r;q)\right)'' = 0.$$ c. The radius of uniform convexity of the function $z \mapsto h_{\nu}^{(s)}(z;q)$ is the smallest positive root of the equation $$(\nu - 1)(\nu - 2)J_{\nu}^{(s)}(\sqrt{r};q) + (4 - 2\nu)\sqrt{r}\left(J_{\nu}^{(s)}(\sqrt{r};q)\right)' + r\left(J_{\nu}^{(s)}(\sqrt{r};q)\right)'' = 0.$$ Proof. The proofs for the cases s=2 and s=3 are almost the same. This is why we are going to present the proof only for the case s=2. a. Let $j_{\nu,n}(q)$ and $j'_{\nu,n}(q)$ be the nth positive roots of the functions $z \mapsto J^{(2)}_{\nu}(z;q)$ and $z \mapsto dJ^{(2)}_{\nu}(z;q)/dz$ , respectively. In [5, p. 979], it was shown that the following equality is valid $$1 + z \frac{\left(f_{\nu}^{(2)}(z;q)\right)''}{\left(f_{\nu}^{(2)}(z;q)\right)'} = 1 - \left(\frac{1}{\nu} - 1\right) \sum_{n \ge 1} \frac{2z^2}{j_{\nu,n}^2(q) - z^2} - \sum_{n \ge 1} \frac{2z^2}{{j'}_{\nu,n}^2(q) - z^2}.$$ In the first step of our proof we consider the case $\nu \geq 1$ . We know that the zeros of Jackson's second and third q-Bessel functions are all real when $\nu > -1$ , according to [18, 20]. Also, it is known from [5, Lemma 9., p. 975] that the zeros of the functions $z \mapsto J_{\nu}^{(s)}(z;q)$ and $z \mapsto dJ_{\nu}^{(s)}(z;q)/dz$ are interlace. Here it is important to mention that the non-negative smallest zero is z=0 for Jackson's second and third q-Bessel functions. By taking $\lambda=1-\frac{1}{\nu}$ in the inequality (1.3) we have $$\Re\left(\frac{2z^2}{{j'}_{\nu,n}^2(q)-z^2}-\left(1-\frac{1}{\nu}\right)\frac{2z^2}{j_{\nu,n}^2(q)-z^2}\right)\leq \left(\frac{2r^2}{{j'}_{\nu,n}^2(q)-r^2}-\left(1-\frac{1}{\nu}\right)\frac{2r^2}{j_{\nu,n}^2(q)-r^2}\right),$$ for $|z| \leq r < j'_{\nu,1}(q) < j_{\nu,1}(q)$ and so, we get that (2.5) $$\Re\left(1+z\frac{\left(f_{\nu}^{(2)}(z;q)\right)''}{\left(f_{\nu}^{(2)}(z;q)\right)'}\right) \ge 1+r\frac{\left(f_{\nu}^{(2)}(r;q)\right)''}{\left(f_{\nu}^{(2)}(r;q)\right)'}.$$ On the other hand, the inequality (1.2) implies that <span id="page-4-0"></span> $$\left| \frac{2z^2}{{j'}_{\nu,n}^2(q) - z^2} - \left(1 - \frac{1}{\nu}\right) \frac{2z^2}{j_{\nu,n}^2(q) - z^2} \right| \le \frac{2r^2}{{j'}_{\nu,n}^2(q) - r^2} - \left(1 - \frac{1}{\nu}\right) \frac{2r^2}{j_{\nu,n}^2(q) - r^2},$$ where $|z| \leq r < j'_{\nu,1}(q) < j_{\nu,1}(q)$ . Therefore, we obtain that (2.6) $$\left| z \frac{\left( f_{\nu}^{(2)}(z;q) \right)''}{\left( f_{\nu}^{(2)}(z;q) \right)'} \right| \le -r \frac{\left( f_{\nu}^{(2)}(r;q) \right)''}{\left( f_{\nu}^{(2)}(r;q) \right)'}.$$ As second step, one can easily show that the inequalities (2.5) and (2.6) hold for $\nu \in (0, 1)$ . Clearly, by considering the inequality (1.4) we can write that <span id="page-4-1"></span> $$\Re\left(\frac{2z^2}{{j'}_{\nu,n}^2(q)-z^2}\right) \le \left|\frac{2z^2}{{j'}_{\nu,n}^2(q)-z^2}\right| \le \frac{2r^2}{{j'}_{\nu,n}^2(q)-r^2}$$ and $$\Re\left(\frac{2z^2}{j_{\nu,n}^2(q)-z^2}\right) \le \left|\frac{2z^2}{j_{\nu,n}^2(q)-z^2}\right| \le \frac{2r^2}{j_{\nu,n}^2(q)-r^2}$$ for $|z| \le r < j'_{\nu,1}(q) < j_{\nu,1}(q)$ . Since $\frac{1}{\nu} - 1 > 0$ , the above last two inequalities imply that the inequalities (2.5) and (2.6) hold true. Consequently, using these two inequalities yields that (2.7) $$\Re\left(1+z\frac{\left(f_{\nu}^{(2)}(z;q)\right)''}{\left(f_{\nu}^{(2)}(z;q)\right)'}\right)-\left|z\frac{\left(f_{\nu}^{(2)}(z;q)\right)''}{\left(f_{\nu}^{(2)}(z;q)\right)'}\right|\geq 1+2r\frac{\left(f_{\nu}^{(2)}(r;q)\right)''}{\left(f_{\nu}^{(2)}(r;q)\right)'}$$ for $|z| \le r < j'_{\nu,1}(q)$ . In (2.7), the equality holds if and only if z = r. Thus, it follows that <span id="page-5-0"></span> $$\inf_{|z| < r} \left[ \Re \left( 1 + z \frac{\left( f_{\nu}^{(2)}(z;q) \right)''}{\left( f_{\nu}^{(2)}(z;q) \right)'} \right) - \left| z \frac{\left( f_{\nu}^{(2)}(z;q) \right)''}{\left( f_{\nu}^{(2)}(z;q) \right)'} \right| \right] = 1 + 2r \frac{\left( f_{\nu}^{(2)}(r;q) \right)''}{\left( f_{\nu}^{(2)}(r;q) \right)'},$$ where $r \in (0, j'_{\nu,1}(q))$ . The mapping $\Phi_{\nu} : (0, j'_{\nu,1}(q)) \mapsto \mathbb{R}$ defined by $$\Phi_{\nu}(r) = 1 + 2r \frac{\left(f_{\nu}^{(2)}(r;q)\right)''}{\left(f_{\nu}^{(2)}(r;q)\right)'} = 1 - 2\sum_{n\geq 1} \left(\frac{2r^2}{j_{\nu,n}^{(2)}(q) - r^2} - \left(1 - \frac{1}{\nu}\right) \frac{2r^2}{j_{\nu,n}^2(q) - r^2}\right)$$ is strictly decreasing since $$\Phi_{\nu}'(r) = -2\sum_{n>1} \left( \frac{4rj_{\nu,n}'^2(q)}{\left(j_{\nu,n}'^2(q) - r^2\right)^2} - \left(1 - \frac{1}{\nu}\right) \frac{4rj_{\nu,n}^2(q)}{\left(j_{\nu,n}^2(q) - r^2\right)^2} \right) < 0$$ for $r \in (0, j'_{\nu,1}(q))$ . Also, we have the following limits $$\lim_{r \searrow 0} \Phi_{\nu}(r) = 1 \text{ and } \lim_{r \nearrow j'_{\nu,1}(q)} \Phi_{\nu}(r) = -\infty.$$ As a result of this, we can say that the equation $$1 + 2r \frac{\left(f_{\nu}^{(2)}(r;q)\right)''}{\left(f_{\nu}^{(2)}(r;q)\right)'} = 0$$ has a unique root $r_0$ in the interval $(0, j'_{\nu,1}(q))$ which is the radius of uniform convexity $r_0 = r^{uc} \left( f_{\nu}^{(2)}(z;q) \right)$ of the function $z \mapsto f_{\nu}^{(2)}(z;q)$ . b. By using logarithmic derivative of the function $z\mapsto dg_{\nu}^{(2)}(z;q)/dz$ which is given by (2.1) we get that <span id="page-5-2"></span>(2.8) $$z \frac{\left(g_{\nu}^{(2)}(z;q)\right)''}{\left(g_{\nu}^{(2)}(z;q)\right)'} = -\sum_{n\geq 1} \frac{2z^2}{\alpha_{\nu,n}^2(q) - z^2}$$ and <span id="page-5-1"></span>(2.9) $$1 + z \frac{\left(g_{\nu}^{(2)}(z;q)\right)''}{\left(g_{\nu}^{(2)}(z;q)\right)'} = 1 - \sum_{n\geq 1} \frac{2z^2}{\alpha_{\nu,n}^2(q) - z^2}.$$ Now, for $|z| \leq r < \alpha_{\nu,1}(q)$ , using the inequality (1.4) in the equalities (2.9) and (2.8), respectively, imply that <span id="page-5-3"></span>(2.10) $$\Re\left(1 + z \frac{\left(g_{\nu}^{(2)}(z;q)\right)''}{\left(g_{\nu}^{(2)}(z;q)\right)'}\right) \ge 1 + r \frac{\left(g_{\nu}^{(2)}(r;q)\right)''}{\left(g_{\nu}^{(2)}(r;q)\right)'}$$ and <span id="page-6-0"></span>(2.11) $$\left| z \frac{\left( g_{\nu}^{(2)}(z;q) \right)''}{\left( g_{\nu}^{(2)}(z;q) \right)'} \right| \le -r \frac{\left( g_{\nu}^{(2)}(r;q) \right)''}{\left( g_{\nu}^{(2)}(r;q) \right)'}.$$ From the inequalities (2.10) and (2.11), we deduce <span id="page-6-1"></span> $$(2.12) \qquad \Re\left(1+z\frac{\left(g_{\nu}^{(2)}(z;q)\right)''}{\left(g_{\nu}^{(2)}(z;q)\right)'}\right) - \left|z\frac{\left(g_{\nu}^{(2)}(z;q)\right)''}{\left(g_{\nu}^{(2)}(z;q)\right)'}\right| \ge 1+2r\frac{\left(g_{\nu}^{(2)}(r;q)\right)''}{\left(g_{\nu}^{(2)}(r;q)\right)'}$$ for $|z| \le r < \alpha_{\nu,1}(q)$ . Equality holds in (2.12) if and only if z = r. As a result, we have $$\inf_{|z| < r} \left[ \Re \left( 1 + z \frac{\left( g_{\nu}^{(2)}(z;q) \right)''}{\left( g_{\nu}^{(2)}(z;q) \right)'} \right) - \left| z \frac{\left( g_{\nu}^{(2)}(z;q) \right)''}{\left( g_{\nu}^{(2)}(z;q) \right)'} \right| \right] = 1 + 2r \frac{\left( g_{\nu}^{(2)}(r;q) \right)''}{\left( g_{\nu}^{(2)}(r;q) \right)'},$$ where $r \in (0, \alpha_{\nu,1}(q))$ . Now consider the function $A_{\nu} : (0, \alpha_{\nu,1}(q)) \mapsto \mathbb{R}$ defined by $$A_{\nu}(r) = 1 + 2r \frac{\left(g_{\nu}^{(2)}(r;q)\right)''}{\left(g_{\nu}^{(2)}(r;q)\right)'} = 1 - \sum_{n \ge 1} \frac{4r^2}{\alpha_{\nu,n}^2(q) - r^2}.$$ The function $A_{\nu}(r)$ is strictly decreasing since $$A'_{\nu}(r) = -\sum_{n \ge 1} \frac{8r\alpha_{\nu,n}^2(q)}{\left(\alpha_{\nu,n}^2(q) - r^2\right)^2} < 0$$ for $r \in (0, \alpha_{\nu,1}(q))$ and also <span id="page-6-2"></span> $$\lim_{r \searrow 0} A_{\nu}(r) = 1 \text{ and } \lim_{r \nearrow \alpha_{\nu,1}(q)} A_{\nu}(r) = -\infty.$$ Therefore, the equation (2.13) $$1 + 2r \frac{\left(g_{\nu}^{(2)}(r;q)\right)''}{\left(g_{\nu}^{(2)}(r;q)\right)'} = 0$$ has a unique root $r_1 \in (0, \alpha_{\nu,1}(q))$ and $r_1 = r^{uc} \left( g_{\nu}^{(2)}(z;q) \right)$ . By using the first and second derivatives of the function $z \mapsto g_{\nu}^{(2)}(z;q)$ , one can easily see that the equation (2.13) is equivalent to <span id="page-6-3"></span> $$(2\nu - 1)(\nu - 1)J_{\nu}^{(2)}(r;q) + (5 - 4\nu)r \left(J_{\nu}^{(2)}(r;q)\right)' + 2r^2 \left(J_{\nu}^{(2)}(r;q)\right)'' = 0.$$ So, the proof is completed. c. The proof of this part can be done similar manner. Logarithmic derivative of the function $z \mapsto dh_{\nu}^{(2)}(z;q)/dz$ which is given by (2.2) implies that (2.14) $$z \frac{\left(h_{\nu}^{(2)}(z;q)\right)''}{\left(h_{\nu}^{(2)}(z;q)\right)'} = -\sum_{n\geq 1} \frac{z}{\beta_{\nu,n}^{2}(q) - z}$$ and <span id="page-7-0"></span>(2.15) $$1 + z \frac{\left(h_{\nu}^{(2)}(z;q)\right)''}{\left(h_{\nu}^{(2)}(z;q)\right)'} = 1 - \sum_{n\geq 1} \frac{z}{\beta_{\nu,n}^{2}(q) - z}.$$ Now, for $|z| \le r < \beta_{\nu,1}^2(q)$ , by using the inequality (1.4) in the equalities (2.15) and (2.14), respectively, we get that <span id="page-7-1"></span>(2.16) $$\Re\left(1+z\frac{\left(h_{\nu}^{(2)}(z;q)\right)''}{\left(h_{\nu}^{(2)}(z;q)\right)'}\right) \ge 1+r\frac{\left(h_{\nu}^{(2)}(r;q)\right)''}{\left(h_{\nu}^{(2)}(r;q)\right)'}$$ and <span id="page-7-2"></span>(2.17) $$\left| z \frac{\left( h_{\nu}^{(2)}(z;q) \right)''}{\left( h_{\nu}^{(2)}(z;q) \right)'} \right| \le -r \frac{\left( h_{\nu}^{(2)}(r;q) \right)''}{\left( h_{\nu}^{(2)}(r;q) \right)'}.$$ Now summarizing the inequalities (2.16) and (2.17), we obtain <span id="page-7-3"></span> $$(2.18) \qquad \Re\left(1+z\frac{\left(h_{\nu}^{(2)}(z;q)\right)''}{\left(h_{\nu}^{(2)}(z;q)\right)'}\right) - \left|z\frac{\left(h_{\nu}^{(2)}(z;q)\right)''}{\left(h_{\nu}^{(2)}(z;q)\right)'}\right| \ge 1+2r\frac{\left(h_{\nu}^{(2)}(r;q)\right)''}{\left(h_{\nu}^{(2)}(r;q)\right)'}$$ for $|z| \leq r < \beta_{\nu,1}^2(q)$ . Equality holds in (2.18) if and only if z = r. Finally, we have $$\inf_{|z| < r} \left[ \Re \left( 1 + z \frac{\left( h_{\nu}^{(2)}(z;q) \right)''}{\left( h_{\nu}^{(2)}(z;q) \right)'} \right) - \left| z \frac{\left( h_{\nu}^{(2)}(z;q) \right)''}{\left( h_{\nu}^{(2)}(z;q) \right)'} \right| \right] = 1 + 2r \frac{\left( h_{\nu}^{(2)}(r;q) \right)''}{\left( h_{\nu}^{(2)}(r;q) \right)'},$$ where $r \in (0, \beta_{\nu,1}^2(q))$ . Now consider the function $B_{\nu} : (0, \beta_{\nu,1}^2(q)) \mapsto \mathbb{R}$ defined by $$B_{\nu}(r) = 1 + 2r \frac{\left(h_{\nu}^{(2)}(r;q)\right)''}{\left(h_{\nu}^{(2)}(r;q)\right)'} = 1 - \sum_{n \ge 1} \frac{2r}{\beta_{\nu,n}^{2}(q) - r}.$$ The function $B_{\nu}(r)$ is strictly decreasing since $$B_{\nu}'(r) = -\sum_{n>1} \frac{2\beta_{\nu,n}^2(q)}{\left(\beta_{\nu,n}^2(q) - r\right)^2} < 0$$ for $r \in (0, \beta_{\nu,1}^2(q))$ and also <span id="page-7-4"></span> $$\lim_{r \searrow 0} B_{\nu}(r) = 1 \text{ and } \lim_{r \nearrow \beta_{\nu,1}^2(q)} B_{\nu}(r) = -\infty.$$ As a result, the equation (2.19) $$1 + 2r \frac{\left(h_{\nu}^{(2)}(r;q)\right)''}{\left(h_{\nu}^{(2)}(r;q)\right)'} = 0$$ has a unique root $r_2 \in (0, \beta_{\nu,1}^2(q))$ and $r_2 = r^{uc} \left( h_{\nu}^{(2)}(z;q) \right)$ . By considering the first and second derivatives of the function $z \mapsto h_{\nu}^{(2)}(z;q)$ , we can easily obtain that the equation (2.19) is equivalent to $$(\nu - 1)(\nu - 2)J_{\nu}^{(2)}(\sqrt{r};q) + (4 - 2\nu)\sqrt{r}\left(J_{\nu}^{(2)}(\sqrt{r};q)\right)' + r\left(J_{\nu}^{(2)}(\sqrt{r};q)\right)'' = 0,$$ which is desired. $\Box$ 2.2. Uniform Convexity of some normalized Wright Functions. In this subsection, we will focus on the function $$\phi(\rho, \beta, z) = \sum_{n \ge 0} \frac{z^n}{n!\Gamma(n\rho + \beta)} \quad (\rho > -1 \text{ and } z, \beta \in \mathbb{C})$$ named after the British mathematician E.M. Wright. It is well known that this function was introduced by him for the first time in the case $\rho > 0$ in connection with his investigations on the asymptotic theory of partitions [30]. From [9, Lemma 1] we know that under the conditions $\rho > 0$ and $\beta > 0$ , the function $z \mapsto \lambda_{\rho,\beta}(z) = \phi(\rho,\beta,-z^2)$ has infinitely many zeros which are all real. Thus, due to the Hadamard factorization theorem, the expression $\lambda_{\rho,\beta}(z)$ can be written as $$\Gamma(\beta)\lambda_{\rho,\beta}(z) = \prod_{n>1} \left(1 - \frac{z^2}{\lambda_{\rho,\beta,n}^2}\right)$$ where $\lambda_{\rho,\beta,n}$ stands for the nth positive zero of the function $\lambda_{\rho,\beta}(z)$ (or the positive real zeros of the function $\Psi_{\rho,\beta}$ ). Moreover, let $\zeta'_{\rho,\beta,n}$ denote the nth positive zero of $\Psi'_{\rho,\beta}$ , where $\Psi_{\rho,\beta}(z) = z^{\beta}\lambda_{\rho,\beta}(z)$ , then the zeros satisfy the chain of inequalities $$\zeta'_{\rho,\beta,1} < \zeta_{\rho,\beta,1} < \zeta'_{\rho,\beta,2} < \zeta_{\rho,\beta,2} < \dots$$ One can easily see that the function $z \mapsto \phi(\rho, \beta, -z^2)$ do not belong to $\mathcal{A}$ , and thus first we perform some natural normalization. We define three functions originating $\phi(\rho, \beta, .)$ : $$f_{\rho,\beta}(z) = \left(z^{\beta}\Gamma(\beta)\phi(\rho,\beta,-z^2)\right)^{\frac{1}{\beta}},$$ $$g_{\rho,\beta}(z) = z\Gamma(\beta)\phi(\rho,\beta,-z^2),$$ $$h_{\rho,\beta}(z) = z\Gamma(\beta)\phi(\rho,\beta,-z).$$ Clearly these functions are contained in the class A. Now, we would like to present our results regarding the uniform convexity of the functions $f_{\rho,\beta}$ , $g_{\rho,\beta}$ and $h_{\rho,\beta}$ .

Theorem 2.2. Let $\rho > 0$ and $\beta > 0$ . a. The radius of uniform convexity of the function $f_{\rho,\beta}$ is the smallest positive root of the equation $$1 + 2r \frac{\Psi_{\rho,\beta}''(r)}{\Psi_{\alpha,\beta}'(r)} + 2\left(\frac{1}{\beta} - 1\right) \frac{r\Psi_{\rho,\beta}'(r)}{\Psi_{\alpha,\beta}(r)} = 0,$$ where $\Psi_{\rho,\beta}(z) = z^{\beta} \lambda_{\rho,\beta}(z)$ . b. The radius of uniform convexity of the function $g_{\rho,\beta}$ is the smallest positive root of the equation $$1 + 2r \frac{g_{\rho,\beta}''(r)}{g_{\rho,\beta}'(r)} = 0.$$ c. The radius of uniform convexity of the function $h_{\rho,\beta}$ is the smallest positive root of the equation $$1 + 2r \frac{h_{\rho,\beta}''(r)}{h_{\rho,\beta}'(r)} = 0.$$